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If $f$ has an anti-derivative in $[a,b]$ does it imply that $f$ is Riemann integrable in $[a,b]$?

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  • $\begingroup$ Integrable in what sense? $\endgroup$ – Chris Eagle Jul 8 '12 at 17:47
  • $\begingroup$ Normally, this is the case (for regular differentiability of functions $f: [a,b] \to \mathbb R$ and both Riemann and Lebesgue integrals). Do you have any specific integral in mind? $\endgroup$ – Johannes Kloos Jul 8 '12 at 17:50
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    $\begingroup$ It is true for Henstock-Kurzweil integral (in this sense H-K integral is better than Riemann and Lebesgue integrals) en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral $\endgroup$ – user8268 Jul 8 '12 at 18:02
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Take $f(x)=\begin{cases} x^2\sin (1/x^2), &x\ne 0, \\ 0, &x=0. \end{cases}\quad$ Then $g=f'$ exists everywhere but is unbounded over $[-1,1]$. $g$ thus has a primitive but is not Riemann integrable.

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The function $f$ need not be Riemann integrable over any non-trivial interval! It can be even arranged that $f$ is bounded. See Volterra's Function.

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